i have a relayset for Arduino where i find these numbers on each relay: 10A-30VDC 10-28VDC

Now, id like to connect that relayset with external devices that run 12V and all the way up to 35A.

I was reading about the logic behind relays and some say that its okay to push in 20A into 10A relay, is just the relay output will not exceed 10A. I also read that the Watts should never exceed, and that's the most important thing.

So taking my relayset 10A x 30VDC = 300 WDC

and my external devices, lets say one of the connectors run on 35A 35A x 12VDC = 420WDC

To my questions:

  1. Is that okay to work with 420W ?
  2. If not, what can happen ?
  3. Is there a way to limit the input to relays so it stays below 300W ?
  4. If i use a transistor (NPN) in the circuit, after the relays, does the relay still take damage from the incoming power ?

I tried to google but is kind of hard to get specific answer to these questions, also id like you to fills out other points (that I'm not mentioning) that i should concern.

  • \$\begingroup\$ Can I ask where you read that logic ? \$\endgroup\$
    – efox29
    May 25, 2015 at 9:42
  • \$\begingroup\$ Do you have a part number for your relay ? \$\endgroup\$
    – efox29
    May 25, 2015 at 9:43
  • \$\begingroup\$ If watts was all that mattered relays would be specified in watts, not volts and amps. \$\endgroup\$ May 25, 2015 at 17:51
  • \$\begingroup\$ what is the actual relay part no.? \$\endgroup\$
    – tcrosley
    May 25, 2015 at 18:21
  • \$\begingroup\$ Thanx for responding!!! Not really sure about which is the part no. but heres a pic of the one i got: kjell.com/image/Product_136661263161658684/full/1 \$\endgroup\$
    – Deko
    May 26, 2015 at 0:15

2 Answers 2


1) Driving 35 amps through contacts rated for 10 amps is a great way to kill the contacts.

2) The contacts will fail either by welding shut or by "burning" so that they can no longer make contact. Which is more likely depends somewhat on your load. If your load is purely resistive or has a turn-on surge (like incandescent lamps or capacitors) then welding is the more likely failure. If the load is inductive (like a motor) then burning the contacts when you try to open them is more likely.

3) In the terms you're asking - no.

4) Yes. (With this exception - if the NPN can be used to interrupt the current flow before the relay is opened, it will prevent burning the contacts. Of course, if you can do this, why are you bothering with a relay?)

5) Whoever said you can push 20A through a 10A relay is giving bad advice, and the concept that "is just the relay output will not exceed 10A" makes no sense at all.

You need a beefier relay.

  • \$\begingroup\$ Explained in a way that made things clear! Thank you ! \$\endgroup\$
    – Deko
    May 26, 2015 at 13:33

I think you need to understand what those numbers mean. 10 A is the maximum rated current. 30 VDC is the maximum recommended operating voltage, the voltage that that the relay is intended to isolate when the contacts are open. 10-28 VDC is the coil operating voltage range, the voltage required for the relay to toggle its state.

  • \$\begingroup\$ thank you for clearing that up for me, had actually no idea at all \$\endgroup\$
    – Deko
    May 26, 2015 at 13:19

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